i, j, that is, if the corresponding entries are equal.
The sum of m x n matrices A = (aij ) and B = (bij ) is the m x n matrix C = (cij ),
whose entries are cij = aij + bij .
The product cA of a matrix A = (aij ) and a number c is the matrix (caij ),
In particular - B = ( - bij ) is the negative of a matrix B = (bij ) and
The product AB (in this order) of matrices Am x n = (aik ) and Bn x p = (bkj ) is the m x p matrix C = (cij ), where
aik bkj .
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a11 b11 + a12 b21 + ··· + a1n bn1 | a11 b12 + ··· + a1n bn2 ··· | ![]() | ||||
| a21 b11 + a22 b21 + ··· + a2n bn1 | a21 b12 + ··· + a2n bn2 ···| : | : | am1 b11 + am2 b21 + ··· + amn bn1 |
| am1 b12 + ··· + amn bn2 ··· | |
Note that the product AB is defined only when the number of the columns of A and the
number of the rows of B coincide.
It is possible that AB is defined while BA is not.
Example 2: Multiplication of matrices
If AB = BA, we say that A and B commute. In this case A and B are n x n matrices and therefore they are square matrices of the same dimension.
Proposition 1 When the operations below are defined, we have
1) A + B = B + A
2) (A + B) + C = A + (B + C)
3) A + O = A
4) A + ( - A) = O
5) (
µ)A =
(µ A)
6) (
+ µ)A =
A + µ A
7)
(A + B) =
A +
B
8) 1· A = A
9) (AB)C = A (BC)
10) A(B + C) = AB + AC
11) (A + B)C = AC + BC
12)
(AB) = (
A)B = A(
B)
13) IA = A
14) AI = A
Proof. Follows from the previous definitions of matrix operations.
Notice! If AB = 0, it does not necessarily follow that A = 0 or B = 0:
Example
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The transpose or the transposed matrix of an m x n matrix A = (aij ) is the n x m matrix AT = (aji ), that is, the rows of AT are the columns of A and the columns of AT are the rows of A.
A on skew symmetric if AT = - A
Proposition 2
1) (AT)T = A
2) (A + B)T = AT + BT
3) (
A)T =
AT
4) (AB)T = BTAT
Problem: Show that a square matrix A can be written as the sum of a symmetric and a skew symmetric matrix: A = ½(A + AT) + ½(A - AT).
In applications (for example, in signal processing) one often needs real symmetric Toeplitz matrices (for example, autocorrelation matrix)
| r (0) | r (1) | r (2) | ··· | r (n - 1) | ![]() | . | ||||||||||||
| r (1) | r (0) | r (1) | ··· | r (n - 2)| r (2) | r (1) | r (0) | ··· | r (n - 3) | : | r (n - 1) | r (n - 2) | r (n - 3) | ··· | r (0)
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If A = (aij ) is a complex matrix then the conjugate matrix of A is
= (
ij ).
A square matrix A is a Hermite matrix (Hermitian),
if AT =
and a skew Hermite matrix (skew Hermitian) if AT = -
.
The
Hermite matrix of A is AH =
T. If A = AH, then A is Hermitian.
The diagonal elements of a Hermite matrix are real, because
aii =
ii .
The diagonal elements of a skew Hermite matrix are pure imaginary or zero, because
aii = -
ii .
Hermiteness generalizes the notion of symmetricness.
Matrix B is the inverse of a matrix A if
Proposition 3. If A has an inverse, the inverse is unique.
Proof. Let B and B' be inverses of a matrix A, that is,
AB = BA = I, and AB' = B'A = I.
But it now follows that B = IB = (B'A)B = B' (AB) = B'I = B'.
If the inverse of A exists, it is denoted by A - 1 and we say that A is regular. If A does not have an inverse, it is singular.
It can be shown that for square matrices A and B we have
Therefore, to prove that B is the inverse of A, it suffices to show that AB = I.
Proposition 4. If A and B are regular and
0, then
1) (A - 1) - 1 = A
2) (
A) - 1 = (1/
)A - 1
3) (AB) - 1 = B - 1A - 1
4) (AT) - 1 = (A - 1)T
Proof. 4): Denote B = (A - 1)T. We have to show that ATB = BAT = I
ATB = AT (A - 1)T = (A - 1A)T = IT = I| BAT = (A - 1)TAT = (AA - 1)T = IT = I | |
A is orthogonal if it is real and AT = A - 1.
If A and B are orthogonal, then so is AB.
Proof. (AB)T = BTAT = B-1A-1 = (AB)-1.
An orthogonal mapping(a matrix) preserves norms!
More generally: m x n matrix U is orthogonal if UTU = I (column orthogonal).
Example 7: Rotation and reflection matrices
A is unitary if
T = A - 1, that is, if
A - 1 = AH. Compare orthogonal vs. unitary.
Often it is useful to partition matrices into smaller ones, called blocks, for example
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Proposition 5. Let matrices A and B be partitioned as follows:
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where the block Aij is a si x tj matrix and Bij is a ui x vj matrix. Then
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3) If p = r, n = m and si = ui , tj = vj
i, j, then
| A + B = |
| C11 | ··· | C1n | ![]() |
| : | : | |
| Cp1 | ··· | Cpn |
where Cij = Aij + Bij
4) If n = r and tj = uj
j, then
| AB = |
| C11 | ··· | C1m | ![]() |
| : | : | |
| Cp1 | ··· | Cpm |
where Cij =
Aik Bkj .
Proof. Straightforward calculation.